Q. Let $u(x, y)$ be the solution of $\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=64$ in the unit disc $\left\{(x, y) \mid x^{2}+y^{2}<1\right\}$ and such that $u$ vanishes on the boundary of the disc. Then $u\left(\frac{1}{4}, \frac{1}{\sqrt{2}}\right)$ is equal to
(1) 7
(2) 16
(3) -7
(4) -16
I tried with Polar form of Laplacian operator to get $$u_{rr}+(1/r)u_r+(1/r^2)u_{\theta \theta}=64$$ and I confused how to deal with the third term $(1/r^2)u_{\theta \theta}$.
Let $F(x,y)=16(x^2+y^2-1)$. Then $F$ vanishes on $x^2+y^2=1$, and $F_{xx}+F_{yy}=64$. $$ F(1/4,1/\sqrt{2})=16(1/16+1/2-1)=1+8-16=-7. $$ The answer is number 3.